# Equal Factors over Equal Intervals

Alignments to Content Standards: F-LE.A.1.a

1. Complete the table below. Is $\Delta x$ a constant? If so, what constant is it? What do you notice about the 3rd column of the table?

$x$ $f(x)=128\cdot \left({\frac {1}{2}}\right)^x$ Successive quotients
0 128 ----
1 64 $\frac{f(1)}{f(0)}=\frac {64}{128} = \frac{1}{2}$
2 $\frac{f(2)}{f(1)}=$
3
4
2. Complete the table below. Is $\Delta x$ a constant? If so, what constant is it? What do you notice about the 3rd column of the table?

3. $x$ $f(x)=128\cdot \left({\frac {1}{2}}\right)^x$ Successive quotients
0 128 ---
2 32 $\frac{f(2)}{f(0)} = \frac {32}{128} = \frac{1}{4}$
4 $\frac{f(4)}{f(2)}=$
6
8
4. Let $f(x)=a \cdot b^x$. Let $x_0$ be any particular $x$-value. Show that if $x_0$ is increased by a constant, $\Delta x$, the successive quotient $$\frac{f(x_0+ \Delta x)}{f(x_0)}$$ is always the same no matter what $x_0$ is.
5. Is b) an example of the result of c)? Explain.

## IM Commentary

This problem assumes that students are familiar with the notation $x_0$ and $\Delta x$. However, the language "successive quotient" may be new. Examples in the third columns of the charts are designed to help students become familiar with this language. Depending on the students's prior exposure to exponential functions and their growth rates, instructors may wish to encourage students to repeat part (b) for a variety of exponential functions and step sizes before proceeding to the most general algebraic setting in part (c).

Two other problems (F-LE Equal Differences over Equal Intervals 1 and F-LE Equal Differences over Equal Intervals 2) illustrate the linear function portion of F-LE.1a.

## Solution

1. $x$ $f(x)=128\cdot \left({\frac {1}{2}}\right)^x$ Successive quotients
0 128 ----
1 64 $\frac {64}{128} = \frac{1}{2}$
2 32 $\frac {32}{64} = \frac{1}{2}$
3 16 $\frac {16}{32} = \frac{1}{2}$
4 8 $\frac {8}{16} = \frac{1}{2}$
$\Delta x$, the change in successive $x$ values, is constant, as its value is 1. The third column in the table always takes the value $\frac{1}{2}$.
2. $x$ $f(x)=128\cdot \left({\frac {1}{2}}\right)^x$ Successive quotients
0 128 ---
2 32 $\frac {32}{128} = \frac{1}{4}$
4 8 $\frac{8}{32} = \frac{1}{4}$
6 2 $\frac{2}{8} = \frac{1}{4}$
8 $\frac{1}{2}$ $\frac{\frac{1}{2}}{2} = \frac{1}{4}$
$\Delta x$, the change in successive $x$ values, is constant, as its value is 2. The third column of the table always takes the value $\frac{1}{4}$.

3. $f(x_0)=a \cdot b^{x_0}$ and $f(x_0+\Delta x)=a \cdot b^{x_0+\Delta x}$. So, the quotient can be rewritten as follows: $$\frac{a \cdot b^{x_0+\Delta x}}{a \cdot b^{x_0}} = \frac{b^{x_0} \cdot b^{ \Delta x}}{b^{x_0}}=b^{ \Delta x}$$ Since both $b$ and $\Delta x$ are constants, the quotient is constant. In particular, the quotient does not depend on the value of $x_0$.
4. In b) the $\Delta x = 2$ and the base, $b$, is $\frac {1}{2}$. The constant successive quotients were $\frac{1}{4}=\frac{1}{2}^{2}=b^{ \Delta x}$. So b) is an example of the result in c).